Conjugate-Beam Method Conjugate-Beam Method

Structure Analysis I Chapter 8 Deflections Introduction

• Calculation of deflections is an important part of structural analysis • Excessive beam deflection can be seen as a mode of failure. – Extensive glass breakage in tall buildings can be attributed to excessive deflections – Large deflections in buildings are unsightly (and unnerving) and can cause cracks in ceilings and walls. – Deflections are limited to prevent undesirable vibrations Beam Deflection

• Bending changes the initially straight longitudinal axis of the beam into a curve that is called the Deflection Curve or Elastic Curve Beam Deflection

• To determine the deflection curve: – Draw shear and moment diagram for the beam – Directly under the moment diagram draw a line for the beam and label all supports – At the supports displacement is zero – Where the moment is negative, the dfldeflect ion curve is concave downward. – Where the moment is positive the deflection curve is concave upward – Where the two curve meet is the Inflection Point

Deflected Shape Example 1 Draw the deflected shape for each of the beams shown Example 2 Draw the deflected shape for each of the frames shown Double Integration Method Elastic‐Beam Theory

• Consider a differential element of a beam subjected to pure bending. • The radius of curvature ρ is measured from the center of curvature to the neutral axis • Since the NA is unstretched, the dx=ρdθ Elastic‐Beam Theory

• Applying Hooke’s law and the Flexure formula, we obtain: 1 M = ρ EI Elastic‐Beam Theory • The product EI is referred to as the flexural rigidity. • Since dx = ρdθ, then M dθ = dx (Slope) EI In most calculus books 1 d 2v / dx2 = ρ 3 [1+ (dv / dx)2 ]2 M d 2v / dx2 = (exact soltilution) EI 3 []1+ ()dv / dx 2 2 d 2v M = dx2 EI The Double Integration Method Relate Moments to Deflections

d 2v M = dx2 EI

dv M(x) Do Not θ(x) = = dx Integration Constants dx ∫ EI(x) Use Boundary Conditions to Evaluate Integration M (x) Constants v(x) = dx 2 ∫∫ EI (x) Assumptions and Limitations

Deflections caused by shearing action negligibly small compared to bending

Deflections are small compared to the cross‐sectional dimensions of the beam

All portions of the beam are acting in the elastic range

Beam is straight prior to the application of loads y L Examples x M = −PL + Px PL x d 2 y P EI = M P 2 d 2 y dx @ x EI = −PL + Px dx2 dy x2 Integrating once EI = −PLx + P + c dx 2 1 2 dy 0 () @ x = 0 = 0 ⇒ EI()0 = −PL 0 + ()P + c ⇒ c = 0 dx 2 1 1 PLx2 x3 Integrating twice EIy = − + P + c 2 6 2 PL ()0 3 @ x = 0 y = 0 ⇒ EI(0)= − (0)2 + P + c ⇒ c = 0 2 6 2 2 PL3 ∆ = @ x = L y = ymax max 3EI PLx2 x3 EIy = − + P 2 6 PL L2 L3 PL3 PL3 EIy = − + P = − ⇒ y = − max 2 6 6 max 3EI y W W M = − (L − x)2 x 2 x d 2 y 2 L WL EI 2 = M 2 dx 2 WL d y W 2 @ x EI = − (L − x) dx2 2

dy W (L − x)3 Integrating once EI = + c dx 2 3 1

dy W (L − 0)3 WL3 @ x = 0 = 0 ⇒ EI()0 = + c ⇒ c = − dx 2 3 1 1 6

dy W WL3 ∴ EI = ()L − x 3 − dx 6 6 W (L − x)4 WL3 Integrating twice EIy = − − x + c 6 4 6 2 W (L − 0)4 WL3 WL4 @ x = 0 y = 0 ⇒ EI(0)= − − (0)+ c ⇒ c = 6 4 6 2 2 24

W WL3 WL4 EIy = − (L − x)4 − x + 24 6 24

Max. occurs @ x = L W L4 WL4 WL4 WL4 EIy = − + = − ⇒ y = − max 6 24 8 max 8EI

WL4 ∆ = max 8EI y Example x x

WL L WL 2 WL x 2 M = x −Wx 2 2 d 2 y WL x2 EI = x −W dx2 2 2 dy WL x2 W x3 Integggrating EI = − + c dx 2 2 2 3 1 L dy Since the beam is symmetric @ x = = 0 2 3 2 dx ⎛ L ⎞ ⎛ L ⎞ ⎜ ⎜ L WL 2 W 2 WL3 @ x = EI()0 = ⎝ ⎠ − ⎝ ⎠ + c ⇒ c = − 2 2 2 2 3 1 1 24 dy WL W WL3 ∴ EI = x2 − x3 − dx 4 6 24 WL x3 W x4 WL3 Integggrating EIy = − − x + c 4 3 6 4 24 2

WL (0)3 W (0)4 WL3 @ x = 0y = 0 ⇒ EI(0)= − − (0)+ c ⇒ c = 0 4 3 6 4 24 2 2

WL W WL3 ∴ EIy = x3 − x4 − x 12 24 24

5WL4 Max. occurs @ x = L /2 EIy = − max 384

5WL4 ∆ = max 384EI y Example x P x

P L/2 L/2 P 2 L P 2 for 0 < x < M = x 2 2 d 2 y P L EI = x for 0 < x < dx2 2 2 dy P x2 Integggrating EI = + c dx 2 2 1 L dy Since the beam is symmetric @ x = = 0 2 2 dx ⎛ L ⎞ ⎜ 2 L P ⎝ 2 ⎠ PL @ x = EI()0 = + c ⇒ c1 = − 2 2 2 1 16 dy P PL2 ∴ EI = x 2 − dx 4 16 P x3 PL2 Integggrating EIy = − x + c 4 3 16 2

P (0)3 PL2 @ x = 0y = 0 ⇒ EI(0)= − (0)+ c ⇒ c = 0 4 3 16 2 2

P PL2 ∴ EIy = x3 − x 12 16

PL 3 Max. occurs @ x = L /2 EIy = − max 48

PL3 ∆ = max 48EI Example Example 5

Moment‐Area Theorems Moment‐Area Theorems

Theorem 1: The change in slope between any two points on the elastic curve equal to the area of the bending moment diagram between these two points, divided by the product EI. dv2 M dv =⇒=θ dx2 EI dx dMθ ⎛⎞ M =⇒=ddxθ ⎜⎟ dx EI⎝⎠ EI B M θ = dx BA ∫ A EI dt= xdθ ⎛⎞M ddxθ = ⎜⎟ ⎝⎠EI BBMM txdxxdx== BA ∫∫ AAEIEI Moment‐Area Theorems

Theorem 2: The vertical distance of point A on a elastic curve from the tangent drawn to the curve at B is equal to the moment of the area under the M/EI diagram between two points (A and B) about point A .

B M txdx= AB ∫ A EI B M txdx= AB ∫ A EI Example 1

Example 2

Example 3

Example 4

Example 5

Example 6

Example 7 M/EI

‐30/EI ‐ ‐20/EI

⎛⎞20 1 ⎛⎞⎛⎞ 10 2 tCB/ ==+⎜⎟−⋅⋅21( )+ ⋅− ⎜⎟⎜⎟ ⋅ 2 ⋅ ⋅ 2 ⎝⎠EI23 ⎝⎠⎝⎠ EI 53.33 =−kN ⋅ m3 =0.00741 rad EI Another Solution Conjugate-Beam Method Conjugate-Beam Method

dV d2 M ==ww dx dx 2 dMθ dvM2 == dx EI dx2 EI Integrating

V== wdx M⎡⎤ wdx dx ∫∫∫⎣⎦ ⎛⎞MM⎡⎤ ⎛⎞ θ = ∫∫∫⎜⎟dx v= ⎢ ⎜⎟ dx⎥ dx ⎝⎠EI⎣ ⎝⎠ EI ⎦ Conjugate-Beam Supports

Example 1 Find the Max. deflection Take E=200Gpa, I=60(106) 562.5 θ = V = − B B' EI 562.5 −14062.5 ∆ = M = (25) = B B' EI EI Example 2 Find the deflection at Point C

C 27 63 −162 ∆ = M = (1) − (3) = C C' EI EI EI Example 3 Find the deflection at Point D 3600 ∆ D = M D' = 720 360 EI EI EI Example 4 Find the Rotation at A

Moment Diagrams and Equations for Maximum Deflection

Example 4 Find the Maximum deflection for the following structure based on The ppgrevious diagrams